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Zinaida [17]
2 years ago
9

The number of students in an elementary school t years after 2002 is given by s(t) = 100 ln(t + 5) students. The yearly cost to

educate one student t years after 2002 can be modeled as c(t) = 1500(1.05t) dollars per student. (a) What are the input units of the function f(t) = s(t) · c(t)?
Mathematics
1 answer:
nikdorinn [45]2 years ago
6 0

Answer:

I'm so srry but I don't know

Step-by-step explanation:

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Merei and Jona buy the $2000 Super PC special package desktop computer. What is the total amount they will pay? Question respons
omeli [17]

Answer:

$2000

Step-by-step explanation:

Given data

We are given that the price of the super PC is $2000

Since they are buying only one of the super PC they will have to pay

$2000

Hence option C is the correct answer

5 0
1 year ago
In circle N, KL ≅ ML. Circle N is shown. Line segments N J, N M, N L, and N K are radii. Lines are drawn to connect each point o
Sergeu [11.5K]

Answer:

arc\ JM=132^o

Step-by-step explanation:

The picture of the question in the attached figure

step 1

we know that

arc\ JM+arc\ ML+arc\ LK+arc\ KJ=360^o ----> by complete circle

substitute the given values

(13x+2)^o+(8x-3)^o+(7x+7)^o+(5x+24)^o=360^o

solve for x

(33x+30)^o=360^o\\33x=360-30\\33x=330\\x=10

step 2

Find the measure of arc JM

arc\ JM=(13x+2)^o

substitute the value of x

arc\ JM=(13(10)+2)^o=132^o

8 0
2 years ago
Read 2 more answers
Identify the GCF of 6x2y2 − 8xy2 + 10xy3. 6x2y2 2x2y 2xy2 6xy2
sattari [20]
First find GCF of the numbers:-

GCF of 2, 6 and 8  = 2
GCF od x^2 x and x  is x
GCF of y^2,  y^2 and y^3  is y^2

so the answer is  2xy^2

The third choice.
3 0
2 years ago
Read 2 more answers
A row of tiny red beads is placed at the beginning and at the end of a 20‐centimeter bookmark.
AlladinOne [14]

Answer:

51 rows

Step-by-step explanation:

Given

Length of bookmark = 20cm

Distance between beads = 4mm

Required

Number of rows of beads

First, the distance between the rows of beads must be converted to cm

if 1mm = 0.1cm

then

4mm = 4*0.1cm

4mm = 0.4 cm

This means that each row of beads is placed at 0.4 cm mark.

The distance between each row follows an arithmetic progression and it can be solved as follows;

T_n = a + (n-1)d

Where Tn = 20cm (The last term)

a = 0 cm (The first term)

d = 0.4cm (The distance between each row of beads)

n = ?? (number of rows)

Solving for n; we have the following;

T_n = a + (n-1)d becomes

20 = 0 + (n-1)0.4

20 =  (n-1)0.4

DIvide both sides by 0.4

\frac{20}{0.4} =  \frac{(n-1)0.4}{0.4}

50 =  (n-1)

50 =  n-1

Add 1 to both sides

50 + 1=  n-1 + 1

n = 51

Hence, the number of rows of beads is 51

4 0
2 years ago
M and N are two events P(M) = 0.60, P(N) = 0.20, and P (M and N) = 0.1.
saul85 [17]

Answer:

.7

Step-by-step explanation:

i just took the test :)

8 0
1 year ago
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